Answer to Question #155083 in Differential Equations for Khan

Question

(z²d²÷dz²)w+3÷16(1+z)w=0

Answer 1

1 STEP : Let's replace the function

$$w(z)=y(z)\cdot z^{1/4}\to\\[0.3cm] \left\{\begin{array}{l} w'_z=y'\cdot z^{1/4}+\displaystyle\frac{1}{4}\cdot y\cdot z^{-3/4}\\[0.3cm] w''_{zz}=y''\cdot z^{1/4}+\displaystyle\frac{2}{4}\cdot y'\cdot z^{-3/4}-\displaystyle\frac{3}{16}\cdot y\cdot z^{-7/4}\\[0.3cm] \end{array}\right.\to\\[0.3cm] \left\{\begin{array}{l} w'_z=y'\cdot z^{1/4}+\displaystyle\frac{y}{4z^{3/4}}\\[0.3cm] w''_{zz}=y''\cdot z^{1/4}+\displaystyle\frac{y'}{2z^{3/4}}-\displaystyle\frac{3y}{16z^{7/4}}\\[0.3cm] \end{array}\right.\to\\[0.3cm] z^2\cdot w''_{zz}+\frac{3}{16}(z+1)w=0\to\\[0.3cm] z^2\cdot\left(y''\cdot z^{1/4}+\frac{y'}{2z^{3/4}}-\frac{3y}{16z^{7/4}}\right)+\frac{3}{16}\cdot(z+1)\cdot y\cdot z^{1/4}=0\\[0.3cm] z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3}{16}\cdot z^{1/4}\cdot y-\frac{3(z+1)}{16}\cdot z^{1/4}\cdot y=0\\[0.3cm] z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3(z+1-1)}{16}\cdot z^{1/4}\cdot y=0\\[0.3cm] \left.z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3}{16}\cdot z^{5/4}\cdot y=0\right|\div\left(z^{5/4}\right)\\[0.3cm] \boxed{z\cdot y''+\frac{y'}{2}-\frac{3}{16}\cdot y=0}\\[0.3cm]$$

2 STEP : Let's change the variable

$$z=x^2\to y'_x=\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}=y'\cdot2x\\[0.3cm] \boxed{y'=\frac{y'_x}{2x}}\\[0.3cm] y''_{xx}=\frac{d}{dx}\left(y'_x\right)=\frac{d}{dx}\left(y'\cdot 2x\right)=2x\cdot\frac{d\left(y'\right)}{dz}\cdot\frac{dz}{dx}+y'\cdot2\\[0.3cm] y''_{xx}=y''\cdot 4x^2+\frac{y'_x}{2x}\cdot2\to\boxed{y''=\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{x}\right)}$$

$$z=x^2\to y'_x=\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}=y'\cdot2x\\[0.3cm] \boxed{y'=\frac{y'_x}{2x}}\\[0.3cm] y''_{xx}=\frac{d}{dx}\left(y'_x\right)=\frac{d}{dx}\left(y'\cdot 2x\right)=2x\cdot\frac{d\left(y'\right)}{dz}\cdot\frac{dz}{dx}+y'\cdot2\\[0.3cm] y''_{xx}=y''\cdot 4x^2+\frac{y'_x}{2x}\cdot2\to\boxed{y''=\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{x}\right)}\\[0.3cm] z\cdot y''+\frac{y'}{2}+\frac{3}{16}\cdot y=0\to\\[0.3cm] x^2\cdot\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{2x}\right)+\frac{1}{2}\cdot\frac{y'_x}{x}+\frac{3}{16}\cdot y=0\\[0.3cm] \left.\frac{y''_{xx}}{4}-\frac{1}{4}\cdot\frac{y'_x}{x}+\frac{1}{4}\cdot\frac{y'_x}{x}+\frac{3}{16}\cdot y=0\right|\cdot(4)\\[0.3cm] \boxed{y''_{xx}+\frac{3}{4}\cdot y=0}\\[0.3cm] \text{The solution will be sought in the form}\quad y(x)=e^{kx}\to\\[0.3cm] y''_{xx}=k^2\cdot e^{kx}\to k^2\cdot e^{kx}+\frac{3}{16}\cdot e^{kx}=0\\[0.3cm] e^{kx}\cdot\left(k^2+\frac{3}{16}\right)=0\to \left[\begin{array}{l} k_1=\sqrt{-\displaystyle\frac{3}{4}}\equiv\displaystyle\frac{i\cdot\sqrt{3}}{2}\\[0.4cm] k_2=-\sqrt{-\displaystyle\frac{3}{4}}\equiv-\displaystyle\frac{i\cdot\sqrt{3}}{2} \end{array}\right.\\[0.3cm]$$

Then, the solution has the form

$$y(x)=A_1\cdot e^{k_1x}+A_2\cdot e^{k_2x}\equiv C_1\cdot\cos\left(\frac{x\sqrt{3}}{2}\right)+C_2\cdot\sin\left(\frac{x\sqrt{3}}{2}\right)\\[0.3cm]$$

It remains to return to the initial variable and function : $z=x^2\to x=\sqrt{z}$ and $w(z)=y(z)\cdot z^{1/4}$.

Conlusion,

$$\boxed{w(z)=z^{1/4}\cdot\left(C_1\cdot\cos\left(\frac{\sqrt{3z}}{2}\right)+C_2\cdot\sin\left(\frac{\sqrt{3z}}{2}\right)\right)}$$

ANSWER

$$w(z)=z^{1/4}\cdot\left(C_1\cdot\cos\left(\frac{\sqrt{3z}}{2}\right)+C_2\cdot\sin\left(\frac{\sqrt{3z}}{2}\right)\right)$$